If $y=\cot x$ show that $\frac{d^{2} y}{d x^{2}}+2 y \frac{d y}{d x}=0$
Formula: -
(i) $\frac{d y}{d x}=y_{1}$ and $\frac{d^{2} y}{d x^{2}}=y_{2}$
(ii) $\frac{\mathrm{d}(\cot x)}{d x}=-\operatorname{cosec}^{2} x$
(iii) $\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{x}^{\mathrm{n}}=\mathrm{nx}^{\mathrm{n}-1}$
(iv) chain rule $\frac{\mathrm{df}}{\mathrm{dx}}=\frac{\mathrm{d}(\text { wou })}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}=\frac{\mathrm{dw}}{\mathrm{ds}} \cdot \frac{\mathrm{ds}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}$
Given: -
$Y=\cot x$
Differentiating w.r.t. $x$
$\frac{d y}{d x}=\frac{d(\cot x)}{d x}$
Using formula (ii)
$\Rightarrow \frac{d y}{d x}=-\operatorname{cosec}^{2} x$
Differentiating w.r.t $\mathrm{x}$
$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-[2 \operatorname{cosecx}(-\operatorname{cosecxcotx})]$
Using formual (iii)
$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=2 \operatorname{cosec}^{2} \mathrm{x} \cot \mathrm{x}$
$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-2 \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \mathrm{y}$
$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=0$
Hence proved.